Question

solve the given initial-value problem$\left(\frac{1}{1+y^2}+cosx-2xy\right)\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{y}\mathbf{(}\mathit{y}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The initial value of the problem is$-x{y}^2+ycosx+ta{n}^{-1}y=1+\frac{\pi }{4}$

Step-by-step solution

Given Information

The given equation is, $\left(\frac{1}{1+y^2}+\cos x-2xy\right)\frac{dy}{dx}=y\left(y+\sin xx\right)$.

Compare the given equation with$Mdx+Ndy=0$

$\begin{array}{l}M(x,y)=-y(y+sinx)=-y^2-ysinx\\ N(x,y)=\frac{1}{1+y^2}+cosx-2xy\end{array}$

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, if$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial t}$

$\begin{array}{l}\frac{\partial M}{\partial y}=-2y-sinx\\ \frac{\partial N}{\partial x}=-sinx-2y\end{array}$

Since, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Therefore, the equation is exact.

Find the solution

Now, consider the equation as, $f(x,y)=-x{y}^2+ycosx+g\left(y\right)$

Integrate $M(x,y)=\frac{\partial f}{\partial x}=-y^2-ysinx$

$\frac{\partial f}{\partial y}=-2xy+cosx+g^{¢}\left(y\right)$

$g^{¢}\left(y\right)=\frac{1}{1+y^2}$

Since $N(x,y)=\frac{\partial f}{\partial y}$then $\begin{array}{l}g\left(y\right)=ta{n}^{-1}y\\ g^{¢}\left(y\right)=\frac{1}{1+y^2}\end{array}$

So the solution is

role="math" $-x{y}^2+ycosx+ta{n}^{-1}y=c-x{y}^2+ycosx+ta{n}^{-1}y=c$

Substitute initial condition

$\begin{array}{c}0+1+\frac{\pi }{4}=c\\ c=1+\frac{\pi }{4}\end{array}$

Thus, the required solution is$-x{y}^2+ycosx+ta{n}^{-1}y=1+\frac{\pi }{4}$